Introduction used. We can also say that if

Introduction

Probabilistic
analysis:-

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This algorithm mostly used in that algorithm that is
complex and difficult to understand. In this algorithm we have to understand
the problem and have to observe the inputs of that algorithm. Those
observations used to design the efficient and attractive algorithm and finish
the complexities of the algorithm.

Here we know that, two cases are occurred, 1st
one is “BEST CASE” and the 2nd one is ”WORST CASE”. One other
situation takes place between the Best and the Worst case, which is “AVERAGE
CASE”. We analyze the Average case very rarely.

So we can say that, probabilistic analysis is used
to understand and describe the Average case behavior.

Moreover we should say that there is no input in
these types of algorithms, we have to estimate the input by our self. Then we
apply the probability to get the efficient and attractive algorithm.  We have to assume the input of any problem
which we have to get from known probability distribution and the output which
is get from assuming the input of any problem should be more efficient and
attractive. By probabilistic, how much we can achieve the target from any
input.

 

Randomized
Algorithm:-

When we have to design the algorithm then we used
the randomized technique.  Sometime we
have to choose the good option, when we have many options for randomization. We
have the coin toss example, coin has two faces (head and tail) and there is a
two possibilities it should be head and tail (Random). So in that situations
randomized algorithm used.

We can also say that if we have the one fixed input
then we have the possibility that there should be a many outputs in the random
way. In this we don’t have need to assume the input.

Randomized algorithm is applicable where the input
size is finite; the reason is that as have to select a randomized or random
number from the input. Suppose we have to select the last number from the input
size, we can select the last number even if we know the size or last value so
that’s why the randomized algorithm is applicable where the input size is fixed
or finite.  

The selection behavior is not only depends on the
input size but also depends on the value produced by Randomized algorithm
(Random number generator).

Suppose we have given the array of 10 elements or
more than 10 elements then the selection of the random number not depends on
the size of array but also depends upon the random number generator which can
either be selected from start or mid or from end.

The
Hiring Problem

There is problem is that, we want to hire a
assistant. And we decided to use a agency. The agency send the unemployed
person one by one. But the company ignore the academic and other ability of
employee and send them one by one for interview. But it is over budget for us
to interview and hire a person because we have paid a few charges for
interviewing a person to the company and also pay the charges to person if we
hire him. If the cost which we pay to company for interview a person is ci
and we interviewed N number of person then the total cost which paid to
company is cin. It is not matter that how many people have hire. But
if the charges that pay to hired assistant is ch then the total cost
that paid to n hired employee is chn. if we hired a assistant and
interview to next person and compare him with current assistance and hire him
if he has best ability as compare to current assistance then it is very costly
for us.

Hire-Assistant(n)

1                   
Best = 0

2                   
for k=1 to n

3                   
interview candidate k

4                   
if candidate k is better than best

5                   
best = k

6                   
hire best

if we analysis it then if the order of person is
increasing order then it will worst case that we hired every candidate which we
interviewed. And it is very costly because we have paid to every candidate who
has hired. So our task is to make such type of algorithm which decrease the
probability to reach worst case every time.

Solution
for Problem

As the worst case reach due to order of candidate.
We can avoid to reach worst case if we take the control over order of candidate
for interview. So for this, we request to agency to provide the list for
interview candidate and everyday call a candidate for interview randomly. For
example if we call RANDOM algorithm with input of candidate list such as ‘a’ to
‘b’ then there is possibility to hire assistant ‘a’ with probability ½ and ‘b’
with probability ½. A call RANDOM(2,6) return 2,3,4,5 or 6 with probability
1/5.  

 

 

 

 

 

 

Birthday
Paradox

In analyzing of algorithms, most of the algorithms
are that in which we know the input values or parameters to analyze of that
algorithm, but sometimes we does not know the input values which we have to use
in analyzing of the algorithm. But in sometimes exact input values for the
algorithms are unknown, in that case we take some assumptions that are the
probabilistic distribution of all the inputs and then analyze the performance
of our algorithm by applying to any random input which we have from the
distribution. And this assumption then lead to design an efficient algorithm.

Now we take the example of birthday paradox to prove
that we have discussed above. The birthday paradox or birthday problem is that
we have a small group of random people and there are 50% chances that two of
them have the same birthday. This is very rare that two persons from a small
group have same birthday. To solve this problem we assume that there are k
people in the room and we take n as total days of year as 365. Let i and j be
the two persons that we compares that they have the same birthday, the
probability that i and j have same birthday is surprisingly few. The
probability that given two persons I and j have same birthday is totally
depends on the random selection of birthdays that it is independent. The
probability that I and j have same birthday is 1/n. It means that if bi
is chosen, then the probability that bj also chose on the same day
is 1/n. If for example there are 23 people in the room then the probability is
at least ½ that 2 people have same birthday.

To solve this problem we define the indicator random
variable Xij that person i and j have the same birthday in which 1
indicates the same birthday and 0 indicates that they have different birthdays.
The probability that I and j have the same birthday is 1/n where n is the total
number of days in a year. Let X be a random variable that will counts the pairs
the persons individually that have the same birthday. So the pair which have
the same birthday from expected number of pairs is at least 1. If for example
we have 28 persons in a room, then we have at least 1 pair that have the same
birthday from given pairs. So by analyzing using both the probability and
indicator random variable the expected number of matching birthdays is 1.

Similarly we have a process in which randomly toss
identical balls into b bin which are independent. We does not know that how
many balls fall into the bins, for this we take the probability. There are
equally likely chances that the tossed ball will fall into the bin or not fall
in the bin. The probability that the balls falls in the bin is 1/b. For example
if we toss n balls into the bin then the probability that the balls fall into
the bin is n/b. If we want that every bin have a ball then approximately blnb
tosses are required for every bin have a ball.